cianfa72's latest activity

I believe I got the point. A trivial bundle is different from a global trivializable bundle. The former is a product while the latter is...

Sorry, R^3 minus (z-axis, unit circle on xy-plane) is an open set of R^3, so it is a (topological) manifold of dim 3 while a regular 2d...

In my experience this definitely takes a great deal of time and effort. Reading something like Spacetime Physics by Taylor and Wheeler...

https://en.wikipedia.org/wiki/Fiber_bundle#Trivial_bundle I'd add that my doubt comes from this picture of ##\mathbb E^1 \times \mathbb...

I've a doubt about the following: take the trivial fiber bundle of the base space ##B## and the fiber ##F## vs the product ##B \times...

Here the equator (i.e. ##x^2 + y^2 = 1, z=w=0##) is a fiber of S^3. So you're saying to take the stereographic projection on ##w=0## of...

Yes, let me say the "process" of sliding the linked pair of fibers occurs along the rays of the stereographic projection from the north...

yes I just wanted to emphasiize that by sliding the linked pair of fibers along the rays emanating from the north pole of the three...

Consider the Hilbert space on ##\mathbb C## of square integrable functions of one variable ##L^2(\mathbb R)##. As claimed in a recent...

Ok, so basically the first step employs the stereographic projection from the north pole of the 3-sphere to R^3 (i.e. ##x_4 = 0##) --...

I guess there are not too many things to do with them. You could analyse their position-momentum uncertainty, you could analyse the 2D...

Ok, so why aren't these Hermite functions employed in quantum physics ? (Or rather, I've never seen them).

The frame in which you are at rest right now is not inertial. You are stationary in this frame--because you are at rest in the frame...

No. By lifting one of them in the fourth dimension out of the 3-sphere in R^4, they can be separated.

Ok, furthermore such Hermite functions (that are defined on the entire R, are measurable w.r.t. the Lebesgue measure on R and the square...